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Chapter 2: Problem 7

Find a linear equation that has the same solution set as the given equation (possibly with some restrictions on the variables.) $$2 x+y=7-3 y$$

Short Answer

Expert verified
The equivalent linear equation is \(2x + 4y = 7\) without restrictions.

Step by step solution

01

Simplify the Given Equation

The given equation is \(2x + y = 7 - 3y\). First, we'll simplify this equation by moving all terms involving \(y\) to one side of the equation. Add \(3y\) to both sides:\[2x + y + 3y = 7\]
02

Combine Like Terms

On the left side of the equation, combine the \(y\) terms:\[2x + 4y = 7\]This is now a simpler linear equation.
03

Consider Possible Restrictions

The original equation might have restrictions when it was transformed. In this case, the transformation from \(2x + y = 7 - 3y\) to \(2x + 4y = 7\) did not introduce division or any operations that could have restricted the variable. Thus, it imposes no restrictions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equation Simplification
Simplifying an equation is like tidying up a room - you're arranging everything so it's easier to see and understand. In the case of linear equations, simplification often involves gathering similar terms together. This process helps make the equation clearer and easier to solve.

In our original equation, \(2x + y = 7 - 3y\), we simplify by moving all the terms with \(y\) to one side. First, add \(3y\) to both sides. This operation helps us gather all the \(y\) terms. Now we're left with:
  • \(2x + y + 3y = 7\)
This step simplifies our equation to \(2x + 4y = 7\). The equation is now less cluttered, showing the relationship between \(x\) and \(y\) more directly.
Solution Set
The solution set of an equation is all the possible values of the variable(s) that make the equation true. For a linear equation like ours, this is typically a line on a graph.

After simplifying the original equation to \(2x + 4y = 7\), every (\(x, y\)) pair that satisfies this equation is part of its solution set. To find particular solutions, you might choose a value for \(x\) and solve for \(y\), or vice versa. Let's say you pick \(x = 1\); plug it into the equation:
  • \(2(1) + 4y = 7\)
Solve to get \(4y = 5\), yielding \(y = \frac{5}{4}\). Hence, (1, \(\frac{5}{4}\)) is in the solution set.
  • Remember, there are infinitely many points in the solution set for a linear equation, each representing a spot on the line that corresponds to the equation.
Variable Restrictions
Variable restrictions usually occur when there's a need to prevent certain values that make the equation undefined, such as dividing by zero.

However, in our linear equation transformation from \(2x + y = 7 - 3y\) to \(2x + 4y = 7\), we didn't introduce any operations like division. This means there are no additional restrictions placed on \(x\) or \(y\) from the simplification process.
  • If any restrictions were present, they would usually appear after a step that involves dividing by a variable, which can lead to issues if that variable equals zero.
Since neither \(x\) nor \(y\) are in any risky positions in our new equation, you can safely assume they can take any real values without impacting the validity of the equation.

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Most popular questions from this chapter

Solve the systems of linear equations over the indicated \(\mathbb{Z}_{p^{*}}\). $$\begin{array}{l} x+2 y=1 \text { over } \mathbb{Z}_{3} \\ x+y=2 \end{array}$$

Suppose the coal and steel industries form an open economy. Every \(\$ 1\) produced by the coal industry requires \(\$ 0.15\) of coal and \(\$ 0.20\) of steel. Every \(\$ 1\) produced by steel requires \(\$ 0.25\) of coal and \(\$ 0.10\) of steel. Suppose that there is an annual outside demand for \(\$ 45\) million of coal and \(\$ 124\) million of steel. (a) How much should each industry produce to satisfy the demands? (b) If the demand for coal decreases by \(\$ 5\) million per year while the demand for steel increases by \$6 million per year, how should the coal and steel industries adjust their production?

For what value(s) of \(k,\) if any, will the systems have (a) no solution, (b) a unique solution, and (c) infinitely many solutions? $$\begin{array}{l} k x+2 y=3 \\ 2 x-4 y=-6 \end{array}$$

Balance the chemical equation for each reaction. \(\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{O}_{2}\) (This reaction takes place when a green plant converts carbon dioxide and water to glucose and oxygen during photosynthesis.

The coefficient matrix is not strictly diagonally dominant, nor can the equations be rearranged to make it sa. However, both the Jacobi and the Gauss- Seidel method converge anyway. Demonstrate that this is true of the Gauss- Seidel method, starting with the zero vector as the initial approximation and obtaining a solution that is accurate to within 0.01. $$\begin{aligned}5 x_{1}-2 x_{2}+3 x_{3} &=-8 \\\x_{1}+4 x_{2}-4 x_{3} &=102 \\\\-2 x_{1}-2 x_{2}+4 x_{3} &=-90\end{aligned}$$
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